Let $A=a\begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix}$ $(a>0)$ and $I=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$ satisfy $A^4+I=\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}$
So please help to tell me about this 2 and 3. Give me some hints or ideas!!
$1$. $A^4=-I\Rightarrow \hat i\rightarrow-\hat i,\hat j\rightarrow-\hat j\Rightarrow A$ is an anti-clockwise rotation matrix with $\theta=\frac{\pi}4$.
Normalize $A$ by choosing $a=\frac1{\sqrt2}$ so that applying $A$ - once results in an anti-clockwise rotation of $\frac{\pi}4$ and $\hat i\rightarrow \frac1{\sqrt 2}(\hat i+\hat j),\hat j\rightarrow \frac1{\sqrt 2}(-\hat i+\hat j)$, - four times results in an anti-clockwise rotation of $\pi$ and $\hat i\rightarrow-\hat i,\hat j\rightarrow -\hat j$. Note that $a=-\frac1{\sqrt2}$ also does the job, but $a>0$ is given (Reason why $A$ rotates anti-clockwise).
$2$. How many such rotations are required to turn $\hat j$ to $\hat i$?
$n$ is the minimum positive integer multiple of $\frac{3\pi/2}{\pi/4}=6\Rightarrow n=6$.
$3.$ $A^{2020}$ is an anti-clockwise rotation matrix that turns $\hat i\rightarrow-\hat i,\hat j\rightarrow-\hat j$ because $2020\cdot \frac{\pi}4=505\pi=2n\pi+\pi$.
$A^{2020}=-I$.
For 2: $A^{2}\left(\begin{array}{c} 0\\ 1 \end{array}\right)=\left(\begin{array}{c} -1\\ 0 \end{array}\right)$, so$$A^6\left(\begin{array}{c} 0\\ 1 \end{array}\right)=-A^2\left(\begin{array}{c} 0\\ 1 \end{array}\right)=\left(\begin{array}{c} 1\\ 0 \end{array}\right).$$Hint for 3: $2020$ is an odd multiple of $4$.